Asked by Jack
The thiosulfate ion (S2O32-) is oxidized by iodine as follows:
2S2O32-(aq) + I2(aq) → S4O62-(aq) + 2I-(aq)
In a certain experiment, 4.41×10-3 mol/L of S2O32- is consumed in the first 11.0 seconds of the reaction.
Calculate the rate of production of iodide ion in mol/L/s
2S2O32-(aq) + I2(aq) → S4O62-(aq) + 2I-(aq)
In a certain experiment, 4.41×10-3 mol/L of S2O32- is consumed in the first 11.0 seconds of the reaction.
Calculate the rate of production of iodide ion in mol/L/s
Answers
Answered by
DrBob222
4.41E-3/11 sec is rate of consumption S2O3^2-. Rate of production I^- is same.
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